External Direct Product Commutativity/Sufficient Condition
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Theorem
Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Let $\circ_1$ and $\circ_2$ be commutative operations.
Then $\circ$ is also a commutative operation.
General Result
Let $\ds \struct {S, \circ} = \prod_{k \mathop = 1}^n S_k$ be the external direct product of the algebraic structures $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$.
If $\circ_1, \ldots, \circ_n$ are all commutative, then so is $\circ$.
Proof
Let $\circ_1$ and $\circ_2$ be commutative operations.
\(\ds \tuple {s_1, t_1} \circ \tuple {s_2, t_2}\) | \(=\) | \(\ds \tuple {s_1 \circ_1 s_2, t_1 \circ_2 t_2}\) | Definition of Operation Induced by Direct Product | |||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {s_2 \circ_1 s_1, t_2 \circ_2 t_1}\) | Definition of Commutative Operation | |||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {s_2, t_2} \circ \tuple {s_1, t_1}\) | Definition of Operation Induced by Direct Product |
Thus $\circ$ is commutative.
$\blacksquare$
Also see
- External Direct Product Associativity
- External Direct Product Identity
- External Direct Product Inverses
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.1: \ 1^\circ$
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.1$